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Sonalidevarajan:
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H = 0.5 m
g = 10 m/s^2
M = 20 Kg
from 3rd equation of motion
v^2 = u^2 + 2gH
it's dropped so initial velocity = 0
v = √2gH = √(2 x 10 x 0.5) = √10 m/s ----(i)
so kinetic energy when it just reaches the ground is
KE = 1/2 x m x v^2
KE = 1/2 x 20 x 10 ,by 1st equation
KE = 100 jolue
i hope it will help you
regards
g = 10 m/s^2
M = 20 Kg
from 3rd equation of motion
v^2 = u^2 + 2gH
it's dropped so initial velocity = 0
v = √2gH = √(2 x 10 x 0.5) = √10 m/s ----(i)
so kinetic energy when it just reaches the ground is
KE = 1/2 x m x v^2
KE = 1/2 x 20 x 10 ,by 1st equation
KE = 100 jolue
i hope it will help you
regards
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